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Difficulty:
35%
(medium)
Question Stats:
70% (01:42) correct
30%
(02:03)
wrong
based on 76
sessions
History
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If K is a multiple of 29, is K∙Y a multiple of 174?
(1) Y is a multiple of 27.
(2) K is divisible by 2 without a remainder.
(1) Y is a multiple of 27.
(2) K is divisible by 2 without a remainder.
22 May 2021, 04:45
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If K is a multiple of 29, is K∙Y a multiple of 174?
Is ky a multiple of 174 = 2*3*29? We know that k is a multiple of 29, hence in order ky to be a multiple of 2*3*29, we need any of them to contain 2 and 3.
(1) Y is a multiple of 27 --> y is a multiple of 3. We need 2. Not sufficient.
(2) K is divisible by 2 without a remainder --> k is a multiple of 2 (even). We need 3. Not sufficient.
(1)+(2) Sufficient.
Answer: C.
Is ky a multiple of 174 = 2*3*29? We know that k is a multiple of 29, hence in order ky to be a multiple of 2*3*29, we need any of them to contain 2 and 3.
(1) Y is a multiple of 27 --> y is a multiple of 3. We need 2. Not sufficient.
(2) K is divisible by 2 without a remainder --> k is a multiple of 2 (even). We need 3. Not sufficient.
(1)+(2) Sufficient.
Answer: C.
General Discussion
14 May 2021, 00:35
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Bunuel
So K=29a.
We are looking at whether K*Y=174b=29*2*3b
(1) Y is a multiple of 27.
So Y=27c. This means KY=29*27ac.
If a*c is even, then yes. Otherwise, no.
(2) K is divisible by 2 without a remainder
So K=29*2d.
Nothing about Y.
If Y is a multiple of 3, yes. Otherwise, no.
Combined
K*Y=29*2*27cd=29*2*3*18cd=174*18cd.
Sufficient
C
